Answer
$4.75\;m^3$
Work Step by Step
According to Archimedes’ principle the buoyant force is equal to the car’s weight
$F_b=W$
or, $ρ_wgV_{sub}=mg$
or, $V_{sub}=\frac{m}{ρ_w}$
Substituting the given values, we obtain
$V_{sub}=\frac{1800}{1000}\;m^3$
or, $V_{sub}=1.8\;m^3$
Therefore, the volume below the water surface with the car floating is $1.8\;m^3$
Total volume of the car is $(5+0.75+0.8)\;m^3=6.55\;m^3$
Therefore, the volume of the car which is about to disappear below the water surface is
$(6.55-1.8)\;m^3=4.75\;m^3$
